course Mth 158
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11:35:37 query 1.1.20 (was 1.1.12). Explain, step by step, how you solved the equation 5y + 6 = -18 - y
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RESPONSE --> 5y + 6 = -18 - y 6y + 6 = -18 6y = -12 y = -2
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11:37:29 ** STUDENT SOLUTION WITH INSTRUCTOR COMMENT: 5y + 6 = 18 - y Subtract 6 from both sides, giving us 5y = 12 - y Add y to both sides, 5y + y = 12 or 6y = 12 divide both sides by 6 y = 2 INSTRUCTOR COMMENT: This is correct for equation 5y + 6 = 18 - y but the equation as I note it is 5y + 6 = -18 - y. If that's the correct equation the solution is found by practically the same steps but you end up with y = -4. **
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RESPONSE --> ??
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11:45:13 query 1.1.44 (was 1.1.30). Explain, step by step, how you solved the equation (2x+1) / 3 + 16 = 3x
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RESPONSE --> ((2x + 1) / 3) + 16 = 3x multiply 3 with 3,16, and 3x (3 and 3 cancel out) 2x + 1+ 48 = 9x 2x + 49 = 9x subtract 2x from 9x 49 = 7x divide 7 = x
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11:45:23 ** STUDENT SOLUTION: (2x + 1) / 3 + 16 = 3x First, multiply both sides of the equation by 3 2x +1 + 48 =9x or 2x + 49 = 9x subtract 2x from both sides. 49 = 7x Divide both sides by 7 7 = x **
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RESPONSE --> correct
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11:53:06 query was 1.1.36. Explain, step by step, how you solved the equation (x+2)(x-3) = (x+3)^2
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RESPONSE --> (x+2)(x-3) = (x+3)^2 distribute x^2 - 3x + 2x - 6 = x^2 + 6x + 9 subtract x^2 -x - 6 = 6x + 9 add x -6 = 7x + 9 subtract 9 -15 = 7x divide 7 -15/7 = x
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00:03:15 ** STUDENT SOLUTION: (x+2)(x+3) = (x+3)^2 First, we use the distributive property to remove the parenthesis and get x^2 - x - 6 = x^2 + 6x + 9 subtract x^2 from both sides, -x - 6 = 6x + 9 Subtract 9 from both sides - x - 6 - 9 = 6x or -x - 15 = 6x add x to both sides -15 = 7x Divide both sides by 7 -15/7 = x **
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RESPONSE --> correct
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00:04:24 query 1.1.52 (was 1.1.48). Explain, step by step, how you solved the equation x / (x^2-9) + 4 / (x+3) = 3 / (x^2-9)/
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RESPONSE --> x / (x^2-9) + 4 / (x+3) = 3 / (x^2-9) x / (x-3)(x+3) + 4/(x+3) = 3/(x-3)(x+3) (x-3)(x+3)*(x/(x-3)(x+3)) + ((x-3)(x+3)* 4/(x+3)) = ((x-3)(x+3)*3/(x+3)(x-3)) x + 4(x-3) = 3 x + 4x -12 = 3 5x - 12 = 3 5x = 15 x = 3
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00:17:45 ** Starting with x / (x^2 -9) + 4 / (x+3) = 3 / (x^2 -9), first factor x^2 - 9 to get x / ( (x-3)(x+3) ) + 4 / (x+3) = 3 / ( (x-3)(x+3) ). Multiply both sides by the common denominator ( (x-3)(x+3) ): ( (x-3)(x+3) ) * x / ( (x-3)(x+3) ) + ( (x-3)(x+3) ) * 4 / (x+3) = ( (x-3)(x+3) ) * 3 / ( (x-3)(x+3) ). Simplify: x + 4(x-3) = 3. Simplify x + 4x - 12 = 3 5x = 15 x = 3. If there is a solution to the original equation it is x = 3. However x = 3 results in denominator 0 when substituted into the original equation, and division by 0 is undefined. So there is no solution to the equation. When you multiplied both sides by x-3, if x = 3 you were multiplying by zero, which invalidated your solution **
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RESPONSE --> correct i think.
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00:18:51 query 1.1.58 (was 1.1.54). Explain, step by step, how you solved the equation (8w + 5) / (10w - 7) = (4w - 3) / (5w + 7)
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RESPONSE --> (8w +5)/(10w - 7) = (4w - 3)/(5w + 7) multiply both sides by 10w-7 and 5w+7 (5w+7)(8w+5) = (10w-7)(4w-3) distribute both sides 40w^2 + 25w + 54w + 35 = 40w^2 - 30w - 28w + 21 subtract 25w from both sides 40w^2 + 79w + 35 = 40w^2 - 58w +21 subtract 21 from both sides 40w^2 + 79w + 14 = 40w^2 - 58w subtract 40w^2 from both sides 79w + 14 = -58w subtract 79 from both sides 14 = -137w divide -137 from both sides w = -14/137
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00:29:59 ** STUDENT SOLUTION: 1) clear the equation of fractions by multiplying both sides by the LCM (10w - 7)(5W + 7) After cancellation the left side reads: (5w+7)(8w + 5) After cancellation the right side reads: (10w - 7)(4w - 3) multiply the factors on each side using the DISTRIBUTIVE LAW Left side becomes: (40w^2) + 81w + 35 Right side becomes: (40w^2) - 58w + 21 3) subtract 40w^2 from both sides add 58w to both sides subtract 35 from both sides Rewrite: 139w = - 14 Now divide both sides by 139 to get w = - (14 / 139) DER**
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RESPONSE --> correct
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00:32:32 query 1.1.70 (was 1.1.78). Explain, step by step, how you solved the equation 1 - a x = b, a <> 0.
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RESPONSE --> 1 - ax = b, a<>0. i think you add -1???? -1 - ax + 1 = b - 1 -ax = b - 1 x = b - 1 / -a x = -b + 1/ a
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00:42:46 ** Start with 1 -ax = b, a <> 0. Adding -1 to both sides we get 1 - ax - 1 = b - 1, which we simplify to get -ax = b - 1. Divide both sides by -a, which gives you x = (b - 1) / (-a). Multiply the right-hand side by -1 / -1 to get x = (-b + 1) / a or x = (1 - b) / a. **
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RESPONSE --> correct
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00:43:06 query 1.1.72. Explain, step by step, how you solved the equation x^3 + 6 x^2 - 7 x = 0 using factoring.
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RESPONSE --> x^3 + 6 x^2 - 7 x = 0 x(x^2 + 6x - 7) = 0 x(x+7)(x-1) = 0 x=0 ; x+7=0 ; x-1=0 x = 0 ; x = -7 ; x = 1
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00:49:14 ** Starting with x^3 + 6 x^2 - 7 x = 0 factor x out of the left-hand side: x(x^2 + 6x - 7) = 0. Factor the trinomial: x ( x+7) ( x - 1) = 0. Then x = 0 or x + 7 = 0 or x - 1 = 0 so x = 0 or x = -7 or x = 1. **
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RESPONSE --> correct
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00:50:55 query 1.1.90 (was 1.2.18). Explain how you set up and solved an equation for the problem. Include your equation and the reasoning you used to develop the equation. Problem (note that this statement is for instructor reference; the full statement was in your text) scores 86, 80, 84, 90, scores to ave B (80) and A (90).
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RESPONSE --> s = score on exam test avg. of 4 tests: 86+80+84+90 = 340 340 / 4 = 85 1/3 * 85 + 2/3 * s = 80 mult. 3 1 * 85 + 2s = 240 85 + 2s = 240 2s = 155 s = 77.5 He needs at least a 77.5 on final exam to get an 80 B for his final grade. 1/3 * 85 + 2/3 *s = 90 85 + 2s =270 2s = 185 s = 92.5 He needs at least a 92.5 on final exam to get a 90 A on his final grade. -----Since the final exam counted for 2/3 of his grade, that meant that 1/3 of it was the test grades given. I just used that and put an equation together. I took the average of his four test grades given which was 85 and worked it in a problem because 85 was 1/3 of his final grade.
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01:17:07 ** This can be solved by trial and error but the only acceptable method for this course, in which we are learning to solve problems by means of equations, is by an equation. Let x be the score you make on the exam. The average of the four tests is easy to find: 4-test average = ( 86 + 80 + 84 + 90 ) / 4 = 340 / 4 = 85. The final grade can be thought of as being made up of 3 parts, 1 part being the test average and 2 parts being the exam grade. We would therefore have final average = (1 * test average + 2 * exam grade) / 3. This gives us the equation final ave = (85 + 2 * x) / 3. If the ave score is to be 80 then we solve (85 + 2 * x) / 3 = 80. Multiplying both sides by 3 we get 85 + 2x = 240. Subtracting 85 from both sides we have 2 x = 240 - 85 = 155 so that x = 155 / 2 = 77.5. We can solve (340 + x) / 5 = 90 in a similar manner. We obtain x = 92.5. Alternative solution: If we add 1/3 of the test average to 2/3 of the final exam grade we get the final average. So (using the fact that the test ave is 85%, as calculated above) our equation would be 1/3 * 85 + 2/3 * x = final ave. For final ave = 80 we get 1/3 * 85 + 2/3 * x = 80. Multiplying both sides by 3 we have 85 + 2 * x = 240. The rest of the solution goes as before and we end up with x = 77.5. Solving 1/3 * 85 + 2/3 * x = 90 we get x = 92.5, as before. **
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RESPONSE --> i think i got it right.
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01:23:34 query 1.1.82 (was 1.1.90). Explain, step by step, how you solved the equation v = g t + v0 for t.
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RESPONSE --> v = gt + v0 for t. v - v0 = gt divide g both sides (v - v0) / g = t
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01:26:44 ** NOTE: v0 stands for v with subscript 0; the whole expression v0 stands for the name of a variable. It doesn't mean v * 0. Starting with v = g t + v0, add -v0 to both sides to get v - v0 = gt. Divide both sides by g to get (v - v0) / g = t }or t = (v - v0) / g. **
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RESPONSE --> correct
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01:27:13 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> This one was very easy. I like problems like this. They are easy to work out.
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01:28:37 **** Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> ok
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